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Problem 2 The switch moves from open to close at t=0. Use your knowledge of first order circuits to find v(t) for t>0. Show all hand calculations below. Attach graph of v(t). t=0 R1 + v+ R2 C

          Problem 2
The switch moves from open to close at t=0. Use your knowledge of first order circuits to find v(t) for
t>0. Show all hand calculations below. Attach graph of v(t).
t=0
R1
+
v+
R2
C
        
Problem 2
The switch moves from open to close at t=0. Use your knowledge of first order circuits to find v(t) for
t>0. Show all hand calculations below. Attach graph of v(t).
t=0
R1
+
v+
R2
C

Added by James R.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Problem 2 The switch moves from open to close at t=0. Use your knowledge of first order circuits to find v(t) for t>0. Show all hand calculations below. Attach graph of v(t). Problem 2 The switch moves from open to close at t=0. Use your knowledge of first order circuits to find v(t) for t>0. Show all hand calculations below. Attach graph of v(t). t=0 R1 WWWW + vo
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Transcript

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00:01 For this question, solution here is, here for the first one taking t less than 0, so by assuming that the switch was closed for the long time, here the resistance value is 3 kω, here the another resistance of 4 kω and this is the resistance of 2 kω.
00:25 Here the voltage supply is applied 12 v.
00:31 So capacitor voltage is equal to 4 plus 2 divided by 4 plus 2 plus 3, the whole multiplied to 12.
00:44 So by solving here we get the capacitor voltage 8 v.
00:49 Now for the second one, after switch is open, so the circuit becomes, here the resistor value 3 kω, here the resistor 4 kω, this is 2.
01:10 Now apply the kirchhoff voltage law, here the voltage supply of 12 v is given in the circuit.
01:18 Here this is the capacitor voltage which is 100 uf.
01:22 The kirchhoff voltage law is applied in this loop, so we get kvl, the kirchhoff voltage law in the loop, it is minus capacitor voltage plus 4 plus 2 multiplied to k multiplied to i of t is equal to 0.
01:44 Here it is equal to minus 1 divided by c, the whole integral of i with respect to t plus 6 kv multiplied to i of t is equal to, from this by rearranging the equation here we get differentiation of i with respect to t is equal to 10 divided by 6 multiplied to i of t.
02:10 Now for the third one, drawing the laplace transform, transform circuit here, here the resistance is 4000 and this is the resistance 2000...
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